AMC12 2011 A

AMC12 2011 A · Q10

AMC12 2011 A · Q10. It mainly tests Probability (basic), Area & perimeter.

A pair of standard $6$-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?

掷一对标准的$6$面骰子一次。掷出的点数之和决定一个圆的直径。该圆的面积的数值小于该圆周长的数值的概率是多少？

(A) \frac{1}{36} \frac{1}{36}

(B) \frac{1}{12} \frac{1}{12}

(D) \frac{1}{4} \frac{1}{4}

(E) \frac{5}{18} \frac{5}{18}

Answer

Correct choice: (B)

正确答案：（B）

Solution

For the circumference to be greater than the area, we must have $\pi d > \pi \left( \frac{d}{2} \right) ^2$, or $d<4$. Now since $d$ is determined by a sum of two dice, the only possibilities for $d$ are thus $2$ and $3$. In order for two dice to sum to $2$, they must both show a value of $1$. The probability of this happening is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. In order for two dice to sum to $3$, one must show a $1$ and the other must show a $2$. Since this can happen in two ways, the probability of this event occurring is $2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$. The sum of these two probabilities now gives the final answer: $\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{\textbf{B}}$

要使周长大于面积，需满足$\pi d > \pi \left( \frac{d}{2} \right) ^2$，即$d<4$。由于$d$由两枚骰子的点数和决定，因此$d$只能为$2$或$3$。两骰和为$2$时必须都为$1$，其概率为$\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$。两骰和为$3$时，一枚为$1$另一枚为$2$，有两种方式，因此概率为$2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$。两者相加得最终答案：$\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{\textbf{B}}$

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